ln 8 e2x

= ++ dx

4 2 4

1

1 ln 8 x

e + e’x dx

=

21

1

= [ex ’ e’x ]ln 8

1

2

63 e 1

= ’+.

16 2 2e

You Try It: Set up, but do not evaluate, the integral for the arc length of the graph

√

of y = sin x on the interval π/4 ¤ x ¤ 3π/4.

Sometimes an arc length problem is more conveniently solved if we think of the

curve as being the graph of x = g(y). Here is an example.

243

CHAPTER 8 Applications of the Integral

EXAMPLE 8.22

Calculate the length of that portion of the graph of the curve 16x 2 = 9y 3

between the points (0, 0) and (6, 4).

SOLUTION

We express the curve as

3

x = y 3/2 , 0 ¤ y ¤ 4.

4

Then dx/dy = 9 y 1/2 . Now, reversing the roles of x and y in ( ), we ¬nd that

8

the requested length is

4 4

1 + [(9/8)y 1/2 ]2 dy = 1 + (81/64)y dy.

0 0

This integral is easily evaluated and we see that it has value [2 · (97)3/2 ’ 128]/

243.

Notice that the last example would have been considerably more dif¬cult (the

integral would have been harder to evaluate) had we expressed the curve in the

form y = f (x).

You Try It: Write the integral that represents the length of a semi-circle and

evaluate it.

8.5.2 SURFACE AREA

Let f (x) be a non-negative function on the interval [a, b]. Imagine rotating the

graph of f about the x-axis. This procedure will generate a surface of revolution,

as shown in Fig. 8.31. We will develop a procedure for determining the area of such

a surface.

y = f (x)

a b

Fig. 8.31

244 CHAPTER 8 Applications of the Integral

We partition the interval [a, b]:

a = x0 ¤ x1 ¤ x2 ¤ · · · ¤ xk’1 ¤ xk = b.

Corresponding to each pair of elements xj ’1 , xj in the partition is a portion of curve,

y = f (x)

a b

Fig. 8.32

as shown in Fig. 8.32. When that piece of curve is rotated about the x-axis, we obtain

a cylindrical surface. Now the area of a true right circular cylinder is 2π · r · h. We

do not have a true cylinder, so we proceed as follows. We may approximate the

radius by f (xj ). And the height of the cylinder can be approximated by the length

of the curve spanning the pair xj ’1 , xj . This length was determined above to be

about

1/2

1 + [f (xj )]2 xj .

Thus the area contribution of this cylindrical increment of our surface is about

1/2

2π · f (xj ) 1 + [f (xj )]2 xj .

See Fig. 8.33. If we sum up the area contribution from each subinterval of the

partition we obtain that the area of our surface of revolution is about

k

1/2

2π · f (xj ) 1 + [f (xj )]2 (—)

xj .

j =1

But this sum is also a Riemann sum for the integral

b

1/2

f (x) 1 + [f (x)]2

2π dx.

a

As the mesh of the partition gets ¬ner, the sum (—) more closely approximates

what we think of as the area of the surface, but it also converges to the integral.

245

CHAPTER 8 Applications of the Integral

,x

Fig. 8.33

We conclude that the integral

b

f (x)(1 + [f (x)]2 )1/2 dx

2π

a

represents the area of the surface of revolution.

EXAMPLE 8.23

Let f (x) = 2x 3 . For 1 ¤ x ¤ 2 we rotate the graph of f about the x-axis.

Calculate the resulting surface area.

SOLUTION

According to our de¬nition, the area is

2

f (x)(1 + [f (x)]2 )1/2 dx

2π

1

2

= 2π 2x 3 (1 + [6x 2 ]2 )1/2 dx

1

2 3

π

= (1 + 36x 4 )1/2 (144x 3 ) dx.

54 2

1

This integral is easily calculated using the u-substitution u = 36x 4 , du =

144x 3 dx. With this substitution the limits of integration become 36 and 576;

the area is thus equal to

576 3 576

π π

(1 + u)1/2 du = (1 + u)3/2

54 2 54 36

36

π

= [(577)3/2 ’ (37)3/2 ]

54

≈ 793.24866.

246 CHAPTER 8 Applications of the Integral

EXAMPLE 8.24

Find the surface area of a right circular cone with base of radius 4 and

height 8.

SOLUTION

It is convenient to think of such a cone as the surface obtained by rotating

the graph of f (x) = x/2, 0 ¤ x ¤ 8, about the x-axis (Fig. 8.34). According

to our de¬nition, the surface area of the cone is

√

8x 58

[1 + (1/2) ] dx = 2π

2 1/2

2π x dx

02 40

√

= 16 5π.

y

x

Fig. 8.34

You Try It: The standard formula for the surface area of a cone is

S = π r h2 + r 2 .

Derive this formula by the method of Example 8.24.

We may also consider the area of a surface obtained by rotating the graph of a

function about the y-axis. We do so by using y as the independent variable. Here

is an example:

EXAMPLE 8.25

Set up, but do not evaluate, the integral for ¬nding the area of the surface

obtained when the graph of f (x) = x 6 , 1 ¤ x ¤ 4, is rotated about the

y-axis.

247

CHAPTER 8 Applications of the Integral

SOLUTION

We think of the curve as the graph of φ(y) = y 1/6 , 1 ¤ y ¤ 4096. Then the

formula for surface area is

4096

1/2

φ(y) 1 + [φ (y)]2

2π dy.

1

Calculating φ (y) and substituting, we ¬nd that the desired surface area is the

value of the integral

4096

y 1/6 1 + (1/6)y ’5/6

2 1/2

2π dy.

1

You Try It: Write the integral that represents the surface area of a hemisphere of

radius one and evaluate it.

8.6 Hydrostatic Pressure

If a liquid sits in a tank, then it exerts force on the side of the tank. This force is

caused by gravity, and the greater the depth of the liquid then the greater the force.

Pascal™s principle asserts that the force exerted by a body of water depends on

depth alone, and is the same in all directions. Thus the force on a point in the side

of the tank is de¬ned to be the depth of the liquid at that point times the density of

the liquid. Naturally, if we want to design tanks which will not burst their seams, it

is important to be able to calculate this force precisely.

Fig. 8.35

Imagine a tank of liquid having density ρ pounds per cubic foot as shown in

Fig. 8.35. We want to calculate the force on one ¬‚at side wall of the tank. Thus

we will use the independent variable h to denote depth, measured down from the

surface of the water, and calculate the force on the wall of the tank between depths

h = a and h = b (Fig. 8.36). We partition the interval [a, b]:

a = h0 ¤ h1 ¤ h2 ¤ · · · ¤ hk’1 ¤ hk = b.

248 CHAPTER 8 Applications of the Integral

Assume that the width of the tank at depth h is w(h). The portion of the wall between

h = hj ’1 and h = hj is then approximated by a rectangle Rj of length w(hj ) and

width h = hj ’ hj ’1 (Fig. 8.37).

h=a

h=b

Fig. 8.36

w(hj)

Y ,h

FL

Fig. 8.37

AM

Now we have the following data:

Area of Rectangle = w(hj ) · h square feet

TE

Depth of Water ≈ hj feet

Density of Liquid = ρ pounds per cubic foot.

It follows that the force exerted on this thin portion of the wall is about

Pj = hj · ρ · w(hj ) · h.

Adding up the force on each Rj gives a total force of

k k

Pj = hj ρ w(hj ) h.

j =1 j =1

But this last expression is a Riemann sum for the integral

b

(—)

hρw(h) dh.

a

EXAMPLE 8.26

A swimming pool is rectangular in shape, with vertical sides. The bottom

of the pool has dimensions 10 feet by 20 feet and the depth of the water

249

CHAPTER 8 Applications of the Integral

is 8 feet. Refer to Fig. 8.38. The pool is full. Calculate the total force on one

of the long sides of the pool.

20

10

8

Fig. 8.38

SOLUTION

We let the independent variable h denote depth, measured vertically down

from the surface of the water. Since the pool is rectangular with vertical sides,

w(h) is constantly equal to 20 (because we are interested in the long side). We

use 62.4 pounds per cubic foot for the density of water. According to (—), the

total force on the long side is

8 8

h · 62.4 · w(h) dh = h · 62.4 · 20 dh = 39936 lbs.

0 0

You Try It: A tank full of water is in the shape of a cube of side 10 feet. How

much force is exerted against one wall of the tank between the depths of 3 feet and

6 feet?

EXAMPLE 8.27

A tank has vertical cross section in the shape of an inverted isosceles tri-

angle with horizontal base, as shown in Fig. 8.39. Notice that the base of the

tank has length 4 feet and the height is 9 feet. The tank is ¬lled with water to

a depth of 5 feet. Water has density 62.4 pounds per cubic foot. Calculate

the total force on one end of the tank.

4 ft

9 ft

5 ft

Fig. 8.39

250 CHAPTER 8 Applications of the Integral

SOLUTION

As shown in Fig. 8.40, at depth h (measured down from the surface of the

water), the tank has width corresponding to the base of an isosceles triangle

similar to the triangle describing the end of the tank. The height of this triangle

is 5 ’ h. Thus we can solve

4

w(h)

=.

5’h 9

We ¬nd that

4

w(h) = (5 ’ h).

9

According to (—), the total force on the side is then

5 4

h · 62.4 · (5 ’ h) dh ≈ 577.778 lbs.

9

0

4

9

5

5_h

Fig. 8.40

EXAMPLE 8.28

An aquarium tank is ¬lled with a mixture of water and algicide to keep the

liquid clear for viewing. The liquid has a density of 50 pounds per cubic foot.

For viewing purposes, a window is located in the side of the tank, with center

20 feet below the surface. The window is in the shape of a square of side

√

4 2 feet with vertical and horizontal diagonals (see Fig. 8.41). What is the

total force on this window?

SOLUTION

As usual, we measure depth downward from the surface with independent

variable h. Of course the square window has diagonal 4 feet. Then the range of

integration will be h = 20’4 = 16 to h = 20+4 = 24. Refer to Fig. 8.42. For

h between 16 and 20, we notice that the right triangle in Fig. 8.42 is isosceles

and hence has base of length h ’ 16. Therefore

w(h) = 2(h ’ 16) = 2h ’ 32.

251

CHAPTER 8 Applications of the Integral

4√2

Fig. 8.41

h = 16